Matter & Motion - Winter 2001

Chemistry Homework Answer Key #7

Chapter 11 - Solubility Equilibria
 

2. a)


b)


c)


d)

22. The amount of solute in the problem is:
 

According to figure 11.7, the temperature at which this concentration corresponds to the equilibrium concentration of AgNO3 is roughly 30° C.

25. The reaction is, TlIO3 (s) « Tl+ (aq) + IO3- (aq)
So, Ksp = [Tl+] [IO3-].

The problem is solved as a typical equilibrium problem, that is, with a table of concentration changes:

 
Tl+
IO3-
initial concentration
0
0
concentration change
x
x
equilibrium conc.
x
x
 
Ksp = x2 = 3.07 ´ 10-6

Since there is a 1:1 molar ratio between the thallium and thallium iodate, 1.75 ´ 10-3 mol of the material will dissolve in one liter of water. The mass of thallium iodate in 100.0 mL of water is given:


 


31. The reaction is: Ag2CrO4 (s) « 2 Ag+ (aq) + CrO42- (aq)

Ksp = [Ag+]2 [CrO42-]

To find the Ksp from a mass of dissolved compound, you must first calculate the equilibrium concentrations of the component ions in solution.
 


 

Substituting these values into the equilibrium expression yields:

Ksp = [Ag+]2 [CrO42-] = (1.55 ´ 10-4)2 (7.78 ´ 10-5) = 1.87 ´ 10-12.
 

35. The reaction quotient, Q, is given by: Q = [Ba2+] [CrO42-].
At 100°C, the value of Q can be obtained once the concentrations of the ions is calculated.


 

Q = 6.2 ´ 10-10.

When this solution is cooled to 25° C, this value of Q exceeds Ksp (2.1 ´ 10-10). Since Q > K, the system will shift to achieve equilibrium; in this case barium chromate will precipitate until the product of the ion concentrations equals Ksp.


41. In this problem, the silver and chromate solutions are mixed to form a single solution. To calculate the equilibrium concentrations of each, it is helpful to first calculate the initial concentrations of these ions in the mixed solution before any precipitate forms. Use the simple dilution relationship, V1c1 = V2c2, where V2 is the volume of the mixed solution.


 

Using these initial concentrations, a typical equilibrium table can be made. The relevant equilibrium reaction is:

Ag2CrO4 (s) « 2 Ag+ (aq) + CrO42- (aq)

Ksp = [Ag+]2 [CrO42-]

 
Ag+
CrO42-
initial concentration
0.0625
0.0225
concentration change
-2x
-x
equilibrium conc.
0.0625 - 2x
0.0225 - x
 
Substituting into Ksp gives,

Ksp = [Ag+]2 [CrO42-] = (0.0625 - 2x)2 (0.0225 - x) = 1.9 ´ 10-12

This turns out to be a messy cubic equation. Since Ksp is so small, we can simplify the problem by saying that this is similar to a limiting reactant problem. So, silver chromate will precipitate from solution until one of the ions is completely consumed. Once the concentration of the excess ion is determined after the reaction, we can use that value to find how much of the other actually remains in solution.

Since there is a 2:1 stoichiometry, the above initial concentrations indicate that the chromate will be consumed before the silver. So, using a table, let x = 0.0225.

 
Ag+
CrO42-
initial concentration
0.0625
0.0225
concentration change
-2 (0.0225)
-0.0225
equilibrium conc.
0.0175
0
So, if all of the chromate reacts, the concentration of silver that remains is 0.0175 M. Using this value in the expression for Ksp yields:

45a) The reaction is, Ni(OH)2 (s) « Ni2+ (aq) + 2 OH- (aq)
So, Ksp = [Ni2+] [OH-]2.

Neglecting the hydroxide ion present due to the autoionization of water gives the following equilibrium table:

 
Ni2+
OH-
initial concentration
0
0
concentration change
x
2x
equilibrium conc.
x
2x
 
Ksp = (x)(2x)2 = 4x3 = 1.6 ´ 10-16

Since x was defined as the equilibrium concentration of Ni2+, this value is equivalent to the molar solubility of Ni(OH)2.

b) When Ni(OH)2 is dissolved in 0.1 M NaOH, there is a common effect at work. So, the equilibrium table has an initial concentration of OH- as 0.1, not zero as above. Moreover, since the amount of Ni(OH)2 that dissolves is so small, we can approximate the final concentration of OH- to be the same as the initial value. So,
 
Ni2+
OH-
initial concentration
0
0.1
concentration change
x
2x
equilibrium conc.
x
0.1 + 2x
 
Ksp = (x)(0.1 + 2x)2 » (x)(0.1)2 = 0.01x = 1.6 ´ 10-16

x = 1.6 ´ 10-14.

Therefore, the molar solubility of Ni(OH)2 drops by about eight orders of magnitude due to the common ion effect.


53. To find out the oxalate concentration required to precipitate the two ions present, the Ksp expressions are used as shown below. Note that the two ionic compounds of interest are MgC2O4 and PbC2O4.

For Mg2+:

Ksp = [Mg2+] [C2O42-]

8.6 ´ 10-5 = (0.1) [C2O42-]

[C2O42-] = 8.6 ´ 10-4 M
 

For Pb2+:

Ksp = [Pb2+] [C2O42-]

2.7 ´ 10-11 = (0.1) [C2O42-]

[C2O42-] = 2.7 ´ 10-11 M

Therefore as oxalate is added, Pb2+ will precipitate out of solution first; no Mg2+ will precipitate until the oxalate concentration reaches 8.6 ´ 10-4 M, at which time both Pb2+ and Mg2+ precipitate. Therefore, to explicitly answer the question, the highest oxalate concentration that can exist in this solution while still having a pure precipitate is 8.6 ´ 10-4 M; the precipitate will be pure PbC2O4.

b) At the point the magnesium begins to precipitate, the lead (II) ion concentration is given by:

Since the initial concentration of lead was 0.1 M, 3.1 ´ 10-5 % of the original lead will still be dissolved when magnesium begins to precipitate. Or, to look at it another way, 99.99997% of the Pb2+ have been removed by this selective precipitation.


89. The reaction is, BaSO4 (s) « Ba2+ (aq) + SO42- (aq)

So, Ksp = [Ba2+] [SO42-].

The problem is solved as a typical equilibrium problem, that is, with a table of concentration changes:

 
Ba2+
SO42-
initial concentration
0
0
concentration change
x
x
equilibrium conc.
x
x
 
Ksp = x2 = 1.1 ´ 10-10

So, the concentration of barium would be 1.0 ´ 10-5 M, or 1.4 mg Ba2+ per liter. Yum.


92. The solubility of CaF2 is described by the equilibrium:

CaF2 (s) « Ca2+ (aq) + 2 F- (aq)

Ksp = [Ca2+] [F-]2

 
The maximum amount of fluoride ion concentration possible when the calcium ion has a concentration is 0.0020 M is given by:


 

Thus, the concentration of fluoride that can be reached before the precipitation of CaF2 is seen is greater than the recommended level of 5 ´ 10-5 M.


Additional Problem:
 

a) NiCl2 (aq) + (NH4)2S (aq) ® NiS (s) + 2 NH4Cl (aq)

precipitate: nickel (II) sulfide
 

b) 3 Mn(NO3)2 (aq) + 2 Na3PO4 (aq) ® Mn3(PO4)2 (s) + 6 NaNO3 (aq)

precipitate: manganese (II) phosphate
 

c) Pb(NO3)2 (aq) + 2 KBr (aq) ® PbBr2 (s) + 2 KNO3 (aq)

precipitate: lead (II) bromide
 

d) Ca(NO3)2 (aq) + 2 KF (aq) ® CaF2 (s) + 2 KNO3 (aq)

precipitate: calcium fluoride
 

e) Ca(NO3)2 (aq) + Na2CO3 (aq) ® CaCO3 (s) + 2 NaNO3 (aq)

precipitate: calcium carbonate